Non-Degenerate Perturbation Theory

Let us now generalize our perturbation analysis to deal with systems possessing more than two energy eigenstates. The energy eigenstates of the unperturbed Hamiltonian, $H_0$ , are denoted

$$H_0\, \vert n\rangle = E_n\, \vert n\rangle,$$ (7.19)

where $n$ runs from 1 to $N$ . The eigenkets $\vert n\rangle$ are orthonormal, and form a complete set. Let us now try to solve the energy eigenvalue problem for the perturbed Hamiltonian:

$$(H_0 + H_1) \,\vert E\rangle = E\,\vert E\rangle.$$ (7.20)

We can express $\vert E\rangle$ as a linear superposition of the unperturbed energy eigenkets,

$$\vert E\rangle = \sum_k \langle k \vert E\rangle \vert k\rangle,$$ (7.21)

where the summation is from $k=1$ to $N$ . Substituting the previous equation into Equation (7.20), and right-multiplying by $\langle m\vert$ , we obtain

$$(E_m + e_{mm} - E)\, \langle m\vert E\rangle + \sum_{k\neq m} e_{mk}\, \langle k\vert E\rangle = 0,$$ (7.22)

where

$$e_{mk} = \langle m \vert\,H_1\,\vert k\rangle.$$ (7.23)

Let us now develop our perturbation expansion. We assume that

$$\frac{\vert e_{mk}\vert}{E_m - E_k} \sim {\cal O}(\epsilon),$$ (7.24)

for all $m\neq k$ , where $\epsilon\ll 1$ is our expansion parameter. We also assume that

$$\frac{\vert e_{mm}\vert}{E_m} \sim {\cal O}(\epsilon),$$ (7.25)

for all $m$ . Let us search for a modified version of the $n$ th unperturbed energy eigenstate for which

$$E= E_n + {\cal O}(\epsilon),$$ (7.26)

and

$$\langle n\vert E\rangle =1,$$ (7.27)

$$\langle m\vert E\rangle \sim {\cal O}(\epsilon),$$ (7.28)

for $m\neq n$ . Suppose that we write out Equation (7.22) for $m\neq n$ , neglecting terms that are ${\cal O}(\epsilon^{\,2})$ according to our expansion scheme. We find that

$$(E_m - E_n) \,\langle m \vert E \rangle + e_{mn} \simeq 0,$$ (7.29)

giving

$$\langle m\vert E\rangle \simeq -\frac{e_{mn}}{E_m - E_n}.$$ (7.30)

Substituting the previous expression into Equation (7.22), evaluated for $m=n$ , and neglecting ${\cal O}(\epsilon^{\,3})$ terms, we obtain

$$(E_n + e_{nn} - E) - \sum_{k\neq n} \frac{\vert e_{nk}\vert^{\,2}} {E_k-E_n} \simeq 0.$$ (7.31)

Thus, the modified $n$ th energy eigenstate possesses the eigenvalue

$$E_n' = E_n + e_{nn} + \sum_{k\neq n} \frac{\vert e_{nk}\vert^{\,2}} {E_n-E_k} + {\cal O}(\epsilon^{\,3}),$$ (7.32)

and the eigenket

$$\vert n\rangle' = \vert n\rangle +\sum_{k\neq n}\frac{e_{kn}}{E_n - E_k}\,\vert k\rangle + {\cal O}(\epsilon^{\,2}).$$ (7.33)

Note that

$$\langle m\vert n\rangle' = \delta_{mn} + \frac{e_{nm}^{\,\ast}}{E... ... {E_n-E_m} + {\cal O}(\epsilon^{\,2}) = \delta_{mn} + {\cal O}(\epsilon^{\,2}).$$ (7.34)

Thus, the modified eigenkets remain orthonormal to ${\cal O}(\epsilon^{\,2})$ .

Note, finally, that if the perturbing Hamiltonian, $H_1$ , commutes with the unperturbed Hamiltonian, $H_0$ , then

$$e_{ml}= e_{mm}\,\delta_{ml},$$ (7.35)

and

$$E_n' =E_n + e_{nn},$$ (7.36)

$$\vert n\rangle' =\vert n\rangle.$$ (7.37)

The previous two equations are exact (i.e., they hold to all orders in $\epsilon$ ).